Uniform convergence of a quotient

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SUMMARY

The discussion centers on proving the uniform convergence of the quotient \(\frac{f_n}{g_n}\) given that \(f_n \to f\) and \(g_n \to g\) uniformly on the closed interval \([a,b]\) with \(|g(x)| > 0\). It is established that \(\frac{1}{g_n}\) is defined for sufficiently large \(n\) due to the uniform convergence of \(g_n\) and the non-vanishing condition of \(g\). The challenge lies in demonstrating that \(\frac{f_n}{g_n} \to \frac{f}{g}\) uniformly, particularly emphasizing that this property does not hold if the interval is open, as the boundedness of the limit functions is crucial for the proof.

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Homework Statement


Let f,g be continuous on a closed bounded interval [a,b] with |g(x)| > 0 for all x in [a,b]. Suppose that f_n \to f and g_n \to g uniformly on [a,b]. Prove that \frac{1}{g_n} is defined for large n and \frac{f_n}{g_n} \to \frac{f}{g} uniformly on [a,b]. Show that this is not true if [a,b] is replaced with (a,b).


Homework Equations





The Attempt at a Solution


The fact that g_n \to g uniformly coupled with |g(x)| > 0 is enough for g_n \neq 0 for large enough n, which means that \frac{1}{g_n} is defined. I'm stuck on the other part. The fact that it is seemingly not true for an open interval domain suggests that I need that the limit functions are bounded, but I've not read anything that says a continuous limit implies continuous sequence elements, even under uniform convergence. The converse is of course true, but I'm not sure about this direction of implication. I'm really just stuck on even where to begin trying to prove that |\frac{f_n}{g_n} - \frac{f}{g}| < \epsilon.

Any help whatsoever is greatly appreciated.
 
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You are going to need more than ##g_n(x)\ne 0## for n large. Can you show there is ##d>0## such that ##g_n(x)\ge d > 0## for large n? You need that to show ##1/g_n(x)## is bounded. That should help.
 

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